#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Simultaneous Equations

When there is a system of two trigonometric equations satisfied simultaneously, the system is called simultaneous equation. Such simultaneous equation problems can be divided into two categories:
(i) Two equations in one ‘unknown’ satisfied simultaneously
(ii) Two equations in two ‘unknowns’ satisfied simultaneously

Applications of simultaneous equations is also an important topic. Simultaneous equations questions are quite tricky and can be mastered only through constant practice. We discuss here both the types of examples of simultaneous equations, but first let’s see the general approach of solving such questions.

1. Firstly, find the values of θ lying between 0 and 2π and satisfying the two given equations separately.

2. Select the value of θ which satisfies both the equations and then generalize it.

Illustration 1: Find the most general values of θ which satisfies the equations sin θ = –1/2 and tan θ = 1/√3.

Solution: As discussed above, first solve both the equations separately.

sin θ = –1/2 ⇒ θ = 7π/6 or 11π/6

and tan θ = 1/√3 ⇒ θ = π/6 or 7π/6 [values between 0 and 2π].

Common value of θ = 7π/6

Hence, the required solution is θ = 2nπ + 7π/6.

Watch this Video for more reference

Illustration 2: If tan (A – B) = 1 and sec (A + B) = 2/√3, find the smallest positive values of A and B and their most general values.

Solution: tan (A – B) = 1 ⇒ A – B = π/4 or 5π/4. …… (1)

Also sec (A + B) = 2/√3 ⇒ A + B = π/4 or 5π/4.

Since A + B > A – B, A + B = 11π/6. …… (2)

Solving (1) and (2), we get

A = 25π/24, B = 19π/24 or A = 37π/24, B = 7π/24.

For the most general values tan (A – B) = 1

⇒ A – B = nπ + π/4 …… (3)

sec(A + B) = 2/√3

⇒ cos(A + B) = √3/2

⇒ A + B = 2mπ + π/6. …… (4)

Solving (3) and (4), we get

A = 1/2 [(2m + n) π + π/4 ± π/6]

B = 1/2 [(2m - n) π - π/4 ± π/6] where m, n ∈ I.

Illustration 3: Solve the following system of equations:

sin (x) + cos (y) = 1.5

sin2 (x) + cos2 (y) =1.25.

Solution: Let's make substitution u = sin(x), v = cos(x), then equation can be rewritten as u +  v = 1.5

u2 + v2 = 1.25.

Express v from first equation: v = 1.5−u.

Now, plug this expression into second equation:

This gives u2 + (1.5−u)2 = 1.25 or

2u2 −3u + 1= 0.

This equation has two roots: u1 = 1, u2 = 0.5.

Now, use equation v = 1.5 − u.

1. v= 1.5 − u= 1.5 − 1 = 0.5;

2. v= 1.5 − u= 1.5 − 0.5 = 1.

Thus, we obtained two pairs of values: u1 = 1, v1 = 0.5 and u2 = 0.5, v2 = 1.

Now, if we return to old variables, we will obtain set of two systems:

{sin (x) =1, cos (x) = 0.5

{sin (x) =0.5, cos (x) = 1

Now we shall solve the first system of equations.

Considering sin x =1 we have x = π/2 + 2πk, k Z;

From equation cos (y) = 0.5 we have that y = ±π/3 + 2πn, n Z.

Therefore, this system has solutions of the form π/2+2πk, y = ± π/3 + 2πn, k, n Z.

Now find solutions of the next system:

From equation sin (x) = 0.5 we have that x = (−1)mπ/6 + πm, m Z;

From equation cos (y) =1 we have that y = 2πl, l Z.

Therefore, this system has solutions of the form x= (−1)m π/6 + πmy = 2πlm, l Z.

Thus, initial system has solutions of the form x= π/2 + 2πk, y = ±π/3 + 2πn, k, n Z.

x= (−1)mπ/6 + πmy = 2πlm, l Z.

Illustration 4: Evaluate the value of

(1 + cos π/8) (1 + cos 3π/8) (1 + cos 5π/8) (1 + cos 7π/8).

Solution: The given expression is

(1 + cos π/8) (1 + cos 3π/8) (1 + cos 5π/8) (1 + cos 7π/8)

= (1 + cos π/8) (1 + cos 3π/8) (1 - cos 3π/8) (1 - cos π/8)

= (1 – cos2π/8) (1 - cos23π/8)

= 1/4 [2 sin π/8 sin 3π/8]2

= ¼ [2 sin π/8 cos π/8]2

= ¼ [sin π/4]2

= 1/4. 1/2 = 1/8

Illustration 5: If cos (α – β) = 1 and cos (α + β) = 1/e, where α, β ∈ [-π, π], then the values of α and β satisfying both the equations is/are

Solution:  It is given in the question that cos (α – β) = 1 and cos (α + β) = 1/e, where α, β ∈ [-π, π].

Now, cos (α – β) = 1

α – β = 0, 2π, -2π

Hence, α – β = 0 (since α, β ∈ [-π, π])

So, α = β

Hence, cos 2α = 1/e

So, the number of solutions of above will be number of points of intersection of the curves y = cos 2α and y = 1/e where α, β ∈ [-π, π]

It is quite clear that there are four solutions corresponding to four points of intersection P1, P2, P3 and P4.

Illustration 6: If k = sin (π/18) sin (5π/18) sin (7π/18), then what is the numerical value of k?

Solution: The value of k is given to be k = sin (π/18) sin (5π/18) sin (7π/18).

Hence, k = sin 10° sin 50° sin 70°

= sin 10° sin (60° - 10°).sin (60° + 10°)

= sin 10° [sin260° - sin210°]

= sin 10° [(√3/2)2 - sin210°]

= sin 10° [3/4 - sin210°]

= 1/4 [3sin 10° - 4sin310°]

= 1/4 x sin (3 x 10°)   (since, sin 3θ = 3sinθ - 4sin3 θ)

`        = 1/4 sin 30° `
`        = 1/8.`

To read more, Buy study materials of Trigonometry comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.